Physics Class 9 Chapter 5 Gravitation questions, answers Mardan board

Now come to KIPS notes for 9th Class of Chapter 5, Gravitation Physics Class 9 Chapter 5 Gravitation questions, answers Mardan board (KPK).

Physics Class 9 Chapter 5 Gravitation questions, answers Mardan board

Q.2) Does the sun exert a larger force on the Earth than that exerted on the sun by the Earth? Explain.

Answer:
No, because according to Newton’s law of Universal Gravitation the force of attraction between two objects is given by;

where ‘m_s‘ is the mass of sun
and ‘m_E‘ is the mass of earth
Since the gravitational force of attraction depends on the product of the masses of the sun and the earth. Hence the force exerted by the sun on earth would be equal in magnitude to the force exerted by earth on the sun.

Q.3) What is the importance of gravitational constant ‘G’? Why is it difficult to calculate?

Answer:
The gravitational constant ‘G’ is very important because Newton’s law of gravitation can not be held true without it. According to Newton the force of attraction is directly proportional to the product of masses and inversely proportional to the square of the distance between their centers.

In order to hold it correct a proportionality constant is needed which is G.

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Gravitational force is very difficult to measure because, its magnitude is very small for ordinary masses available in the laboratory at ordinary distances. Therefore it is very difficult to calculate the value of G.

Q.4) If Earth somehow expanded to a larger radius, with no change in mass, how would your weight be affected? How would it be affected if Earth instead shrunk?  

Answer:
The value of g is given by;

If the radius of the earth expands to a larger radius, keeping everything else same, the value of ‘g‘ will be reduced. As w = mg, hence our weight will be reduced.
In the opposite scenario, if earth shrinks, it means the radius of the earth will be reduced. Or in other words, we will be even closer to the center of the earth due to which ‘g‘ increases and hence our weight will also increase.

Q.5) What would happen to your weight on earth if the mass of the earth doubled, but its radius stayed the same?

Answer:
The value of g is given by;

If the mass of the earth is doubled, the value of ‘g’ will be doubled.
As                                                      weight = w = mg
This means the weight would also get doubled. 

Q.6) Why lighter and heavier objects fall at the same rate toward the earth?

Answer:
The gravitational field strength is given by;

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where ‘m_e‘is the mass of the earth
and ‘r_e‘ is the radius of the earth
This shows that ‘g‘ does not depend on the mass of the object. This means that light and heavy bodies should fall towards the centre of the earth with the same acceleration. 

Q.7) The value of ‘g’ changes with location on earth, however, we take the same value of ‘g’ as 9.8 ms^-2 for ordinary calculations. Why?

Answer:
Although the value of ‘g’ varies with location on earth but this variation is small at different places on the surface so we neglect the variation to avoid complications and take 9.8 ms^-2 as a standard.
It is given as;

For example; if we calculate the value of ‘g’ at different altitudes i.e. h = 0, 10 km, 20 km, it can be found that the variation is small;
Here G = 6.67 x 10^-11 Nm²kg^-2
m_e = 6 x 10²⁴ kg
r_e = 6.4 x 10⁶ m

From the above example, the argument that the variation is small is proved.

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Q.8) Moon is attracted by the earth, why it does not fall on earth?

Answer:
As we know that, moon is a natural satellite of the earth. It orbits around the earth with some tangential speed ‘v’. Though the earth attracts the moon towards its center, its tangential speed prevents it from falling towards the earth. The presence of gravitational force between earth and moon provides the necessary centripetal force to the moon due to which it revolves around the earth and does not fall on earth.

Q.9) Why for same height larger and smaller satellites must have same orbital speeds?

Answer:
The orbital speed of satellite is given by the formula;

i.e. the orbital speed depends upon the mass of the earth ‘m_e‘ (the planet around which the satellite is revolving), the radius of the earth and the height from the surface of the earth. It does not depend on the mass or radius of the object orbiting around the planet. Hence the orbital speed would be same for larger and smaller satellites.